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\author{学号 \underline{\hspace{4cm}} \hspace{1cm} 姓名 \underline{\hspace{4cm}} }
\title{复变函数测验3解答}
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\date{2024 年 5 月 20 日}
%\date{March 9, 2021}

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\begin{document}

\maketitle

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\begin{enumerate}

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\item  %Problem 1 第75页例子3.5
设积分路径 $C$ 是连接点 $1$ 和点 $i$ 的直线段，使用参数方程法，计算积分 $$\int_C (x+y+ix^2+iy^2)dz. $$

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{\color{red}解答：

\begin{center}
%\begin{figure}
\includegraphics[height=4cm, width=6cm]{ca-quiz-03-prob-01.png} 
%\end{figure}
\end{center}

设积分路径的参数方程为 $x=t,y=1-t, 0\le t\le 1$. 则 $z=x+iy=t+(1-t)i=i+(1-i)t$. 代入积分可得
\begin{eqnarray*}
\int_C (x+y+ix^2+iy^2)dz &=& \int_{0}^{1} [t+1-t+it^2+i(1-t)^2](1-i)dt \\ 
&=& \int_{0}^{1} (1-i)(1+i-2it+2it^2)dt \\ 
&=& (1-i)\left[ (1+i)t-it^2+\frac{2}{3}it^3 \right] \Big{|}_{0}^{1} \\
&=& (1-i)(1+i-i+\frac{2}{3}i) \\
&=& \frac{5}{3}-\frac{1}{3}i. 
\end{eqnarray*}


}

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\item  %Problem 2 第83页例子3.8
求出 $f(z)=z\sin(z)$ 的原函数，并使用原函数计算积分 
$$\int_{-\pi i}^{\pi i} z\sin (z) dz. $$

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{\color{red}解答：函数 $f(z)=z\sin(z)$ 在复平面上解析，一个原函数为 
$F(z)=\sin(z)-z\cos(z)$. 因此
\begin{eqnarray*}
\int_{-\pi i}^{\pi i} z\sin (z) dz &=& [\sin(z)-z\cos(z)] \mid_{-\pi i}^{\pi i} \\ 
&=& [\sin(\pi i)-\pi i\cos(\pi i)] - [\sin(-\pi i)+\pi i\cos(-\pi i)] \\ 
&=& 2\sin(\pi i) -2\pi i \cos(\pi i) \\
&=& 2\frac{e^{i\pi i} - e^{-i \pi i}}{2i} - 2\pi i\frac{e^{i\pi i} + e^{-i\pi i}}{2} \\ 
&=& -i(e^{-\pi}-e^{\pi}) -\pi i (e^{-\pi}+e^{\pi}) \\
&=& (e^{\pi}-\pi e^{\pi} - e^{-\pi} - \pi e^{-\pi} )i.
\end{eqnarray*}
}

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\item  %Problem 3 第86页例子3.12
设 $C$ 为包含圆周 $|z|=5$ 在内的正向简单闭曲线，使用复周线上的柯西积分定理，
计算积分 $$\int_C \frac{2z+3}{z^2+3z}dz. $$


\vspace{0.2cm}

{\color{red}解答：

\begin{center}
%\begin{figure}
\includegraphics[height=4cm, width=6cm]{ca-quiz-03-prob-03.png}
%\caption{第3题}
%\end{figure}
\end{center}

先将分式函数化简，可得 
$$\frac{2z+3}{z^2+3z} = \frac{1}{z} + \frac{1}{z+3}. $$
记 $C_1,C_2$ 分别是以 $0,-3$ 为中心，半径为1的正向圆周，则有
\begin{eqnarray*}
\int_C \frac{2z+3}{z^2+3z}dz = \int_{C_1} \frac{1}{z}dz + \int_{C_2} \frac{1}{z+3}dz 
= 2\pi i + 2\pi i = 4\pi i. 
\end{eqnarray*}

}

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\item  %Problem 4 第88页例子3.15 
设 $C$ 为圆周 $| z+2i |=1$. 使用柯西积分公式，计算 $$\int_C \frac{dz}{z^2(z^2+4)}.$$

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{\color{red}解答：
\begin{center}
%\begin{figure}
\includegraphics[height=6cm, width=10cm]{ca-quiz-03-prob-04.png}
%\caption{第4题}
%\end{figure}
\end{center}

函数 $f(z)=\frac{1}{z^2(z^2+4)} = \frac{1}{z^2(z-2i)(z+2i)}$ 在复平面内有三个奇点 $0,2i,-2i$. 
函数 $g(z)=\frac{1}{z^2(z-2i)}$ 在圆盘 $|z+2i|\le 1$ 上解析，所以由柯西积分公式可得
\begin{eqnarray*}
\int_C \frac{dz}{z^2(z^2+4)} = \int_C \frac{g(z)}{z+2i}dz 
= 2\pi i g(-2i) = 2\pi i \frac{1}{(-2i)^2(-2i-2i)} = \frac{\pi}{8}. 
\end{eqnarray*}

}

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\item  %Problem 5 第91页例子3.17
使用解析函数的导函数的柯西积分公式，计算积分 $$\int_{|z|=2} \frac{\sin(z)dz}{(z+i)^4}. $$

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{\color{red}解答：

\begin{center}
%\begin{figure}
\includegraphics[height=4cm, width=6cm]{ca-quiz-03-prob-05.png}
%\caption{第5题}
%\end{figure}
\end{center}

函数 $f(z)=\sin(z)$ 在复平面上解析，积分路径 $|z|=2$ 将 $z_0=-i$ 围在内部，所以可以使用导函数的柯西积分公式
\begin{eqnarray*}
f^{(n)}(z_0) = \frac{n!}{2\pi i} \int_C \frac{f(z)}{(z-z_0)^{n+1}}dz. 
\end{eqnarray*}
代入 $f(z)=\sin(z)$, $z_0=-i$ 与 $n=3$, 可得 
\begin{eqnarray*}
\int_{|z|=2} \frac{\sin(z)dz}{(z+i)^4} = \frac{2\pi i}{3!}f^{(3)}(z_0) = \frac{2\pi i}{6} [-\cos(z)]\mid_{z=-i} 
= \frac{\pi i}{3}\cos(-i) = \frac{\pi i}{3} \frac{e^{-ii}+e^{ii}}{2} 
= \frac{\pi i}{6}(e+e^{-1}). 
\end{eqnarray*}

}

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\item  %Problem 6 103页习题12
设函数 $f(z)$ 定义如下，使用柯西积分公式，计算 $f(i)$ 与 $f'(i)$.  
 $$f(z) = \int_{|z|=2} \frac{\zeta^2+3\zeta+4}{\zeta -z}d\zeta. $$ 

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{\color{red}解答：

\begin{center}
%\begin{figure}
\includegraphics[height=4cm, width=6cm]{ca-quiz-03-prob-06.png}
%\caption{第6题}
%\end{figure}
\end{center}

函数 $g(z)=z^2+3z+4$ 在复平面上解析。写出柯西积分公式如下，
\begin{eqnarray*}
g(z) = \frac{1}{2\pi i} \int_C \frac{g(\zeta)}{\zeta-z}d\zeta,  
%g'(z) &=& \frac{1}{2\pi i} \int_C \frac{g(\zeta)}{(\zeta-z)^2}d\zeta. 
\end{eqnarray*}
其中 $|z|<2$. 与题目 $f(z)$ 的定义比较可得 $f(z)=2\pi i g(z)$. 代入 $z=i$ 可得 $$f(i)=2\pi i g(i)=2\pi i (i^2+3i+4)=6\pi (i-1). $$
另一方面，对 $f(z)=2\pi i g(z)$ 求导并代入 $z=i$ 可得
\begin{eqnarray*}
f'(z) &=& 2\pi i g'(z)=2\pi i (2z+3), \\ 
f'(i) &=& 2\pi i (2i+3) = -4\pi +6\pi i. 
\end{eqnarray*}

}

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\item  %Problem 7 第97页例子3.21 
验证 $u=\frac{y}{x^2+y^2}$ 是调和函数，并求以 $u$ 为实部的解析函数 $f(z)$ 使得 $f(1+i)=1-i$. 

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{\color{red}解答：
根据调和函数的定义，要验证 $\Delta u=0$. 计算偏导数可得 
\begin{eqnarray*}
u_x &=& \frac{(0)(x^2+y^2)-y(2x)}{(x^2+y^2)^2} = \frac{-2xy}{(x^2+y^2)^2}, \\ 
u_y &=& \frac{(1)(x^2+y^2)-(y)(2y)}{(x^2+y^2)^2} = \frac{x^2-y^2}{(x^2+y^2)^2}, \\ 
u_{xx} &=& \frac{(-2y)(x^2+y^2)^2-(-2xy)2(x^2+y^2)(2x)}{(x^2+y^2)^4} = \frac{6x^2y-2y^3}{(x^2+y^2)^3}, \\ 
u_{yy} &=& \frac{(-2y)(x^2+y^2)^2-(x^2-y^2)2(x^2+y^2)(2y)}{(x^2+y^2)^4} = \frac{-6x^2y+2y^3}{(x^2+y^2)^3}.
\end{eqnarray*}
因此可得 $\Delta u = u_{xx} + u_{yy} = 0$. 因此 $u(x,y)$ 是调和函数。 

函数 $\frac{1}{z}$ 在去掉原点的复平面上是解析函数，注意到
$$\frac{1}{z}=\frac{\bar{z}}{z\bar{z}} = \frac{x-iy}{x^2+y^2} = -i\left( \frac{y}{x^2+y^2} +i \frac{x}{x^2+y^2} \right), $$
因此所求解析函数为 
$$f(z)=\frac{y}{x^2+y^2} +i \frac{x}{x^2+y^2} +c = \frac{i}{z} + c, $$
其中 $c$ 是任意复数常数。代入条件 $f(1+i)=1-i$, 可得
$\frac{i}{1+i} + c = 1-i$, 求得 $c = \frac{1}{2} - \frac{3}{2}i$. 
因此所求函数为 $$ f(z) = \frac{i}{z} +\frac{1}{2} - \frac{3}{2}i.  $$
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\item  %Problem 8 第106页习题15
%设函数 $f(z)$ 在区域 $D$ 内解析，证明 
%$$\frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} = 4 \frac{\partial^2 f(z)}{\partial z \partial \bar{z}}. $$
设 $f(z)=u+iv$ 是解析函数，设 $u+v=(x-y)(x^2+4xy+y^2)-2(x+y)$. 求 $f(z)$. 

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{\color{red}解答：
将题目中的等式分别对 $x,y$ 求偏导数，可得
\begin{eqnarray}
u_x + v_x &=& (1)(x^2+4xy+y^2) + (x-y)(2x+4y) - 2 = 3x^2+6xy-3y^2-2, \label{eq1}\\ 
u_y + v_y &=& (-1)(x^2+4xy+y^2) + (x-y)(4x+2y) -2 = 3x^2-6xy-3y^2-2 \label{eq2}. 
\end{eqnarray}
因为 $f(z)$ 是解析函数，所以有柯西-黎曼方程 $u_x = v_y$ 与 $u_y = -v_x$. 代入上面第二个等式(\ref{eq2})可得
\begin{eqnarray}
u_x - v_x = u_y+v_y =  3x^2-6xy-3y^2-2. \label{eq3}
\end{eqnarray}
将(\ref{eq1})与(\ref{eq3})联合可得
\begin{eqnarray*}
u_x &=& 3x^2 -3y^2-2, \\
v_x &=& 6xy.  
\end{eqnarray*}
积分可得 
\begin{eqnarray*}
u &=& x^3-3xy^2-2x+\varphi(y), \\ 
v &=& 3x^2y + \psi(y). 
\end{eqnarray*}
其中 $\varphi(y),\psi(y)$ 是待定实值函数。写出 $f(z)=u+iv$ 并与 $z^3-2z$ 比较，可得 
\begin{eqnarray*}
f(z) = u+iv &=& x^3-3xy^2-2x+\varphi(y) + 3x^2yi + \psi(y)i, \\ 
 g(z) = z^3-2z &=& (x+yi)^3-2(x+yi) = x^3+3x^2yi - 3xy^2-y^3i -2x-2yi, \\ 
 f(z)-g(z) &=& \varphi(y) + \psi(y)i +y^3i +2yi. 
\end{eqnarray*}
因为 $f(z)-g(z)$ 是解析函数，但不含 $x$, 因此为复数常数。因此 $\varphi(y)=a$, $\psi(y)=-y^3-2y+b$, 其中 $a,b$为实数常数。 代入题目有关 $u+v$ 的条件，可得 $a+b=0$. 因此所求 $f(z)=z^3-2z+a-ai$, 其中 $a$ 是任意实数。

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